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suss

  • 2 years ago

Xlogy-logz.ylogz-logx.zlogx-logy=1 how to prove this??

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  1. suss
    • 2 years ago
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    it was supposed to x to the power logy-logz y to the power logz-logx z to the power logx-logy

  2. hartnn
    • 2 years ago
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    \(\huge x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}\) like this ?

  3. suss
    • 2 years ago
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    ah thank god..yes like that

  4. hartnn
    • 2 years ago
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    consider \(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}]\) can u simplify that ?

  5. hartnn
    • 2 years ago
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    use, \(\log AB=\log A+\log B \\ \log A^{B}=B\log A\)

  6. hartnn
    • 2 years ago
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    you should get that =0 = ln 1

  7. suss
    • 2 years ago
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    i dont understand y u multiplied the whole by log????

  8. hartnn
    • 2 years ago
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    to be able to simplify

  9. hartnn
    • 2 years ago
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    else, nothing can't be really done to that expression

  10. suss
    • 2 years ago
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    i have done that problem in this way \[x ^{\log y/z}y ^{\log z/x}z ^{\log x/y}\]

  11. suss
    • 2 years ago
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    and i cudnt figure out the next step..... :(

  12. hartnn
    • 2 years ago
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    i know, i tried that also....since i could not proceed, i thought of taking log....

  13. suss
    • 2 years ago
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    if u dont mind,can u show me the simplification process...??? :)

  14. hartnn
    • 2 years ago
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    \(\huge \log[x^{\log y-\log z}y^{\log z-\log x}z^{\log x-\log y}] \\\large =\log[x^{\log y-\log z}]+\log[y^{\log z-\log x}]+\log[z^{\log x-\log y}]\) got this ?

  15. hartnn
    • 2 years ago
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    now use \( \\ \log A^{B}=B\log A\)

  16. hartnn
    • 2 years ago
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    @suss

  17. suss
    • 2 years ago
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    oh sorry man i jus lost connection leeme check it out

  18. suss
    • 2 years ago
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    y did u keep the + sign ???

  19. suss
    • 2 years ago
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    @hartnn

  20. hartnn
    • 2 years ago
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    because \(\log AB=\log A+\log B \\ \)

  21. suss
    • 2 years ago
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    ah finally got it .....thx a ton man

  22. hartnn
    • 2 years ago
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    sure ? u getting that = 0 ? then write 0=log 1

  23. suss
    • 2 years ago
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    yes 100%..i hav to use formula of log ab bak ther ......i really got it man

  24. hartnn
    • 2 years ago
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    good :) welcome ^_^

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